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  • Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (i) maths textbook solution

Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (i) maths textbook solution

Answers (1)

Answer:

\begin{aligned} x=\frac{1-2 k}{3}, y=k \\ \end{aligned}

Given:

\begin{aligned} 6 x+4 y=2,\: \: 9 x+6 y=3 \end{aligned}

Hint:

 A system of two linear equations can have one solution, an infinite number of solution, if a system has no solution it’s called inconsistent

Solution:

\begin{aligned} &6 x+4 y=2 \; \; \; \; \; \; \;....(i) \\ &9 x+6 y=3\; \; \; \; \; \; \;....(ii) \end{aligned}

        \begin{aligned} &A X=B\\ &\text { Here, }\\ &A=\left[\begin{array}{ll} 6 & 4 \\ 9 & 6 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 2 \\ 3 \end{array}\right]\\ &\left[\begin{array}{ll} 6 & 4 \\ 9 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{aligned}

        \begin{aligned} |A| &=\left|\begin{array}{ll} 6 & 4 \\ 9 & 6 \end{array}\right| \\ &=36-36 \\ |A| &=0 \end{aligned}

So, A  is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because

        (\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0

        Let\; C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,

        \begin{aligned} &C_{11}=6, C_{12}=-9, C_{21}=-4, C_{22}=6 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 6 & -9 \\ -4 & 6 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 6 & -4 \\ -9 & 6 \end{array}\right] \end{aligned}

        \begin{aligned} (\text { adjA }) B &=\left[\begin{array}{cc} 6 & -4 \\ -9 & 6 \end{array}\right]\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \\ &=\left[\begin{array}{c} 12-12 \\ -18+18 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \end{aligned}

If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.

Thus, AX = B has infinitely many solutions

Substituting y = k in eqn (i), We get

        \begin{aligned} &6 x+4 k=2 \\ &6 x=2-4 k \\ &x=\frac{2-4 k}{6} \\ &x=\frac{1-2 k}{3} \\ &x=\frac{1-2 k}{3} \text { and } y=k \end{aligned}

The values of x and y satisfy the third equation.

        \begin{aligned} Thus\; x=\frac{1-2 k}{3} \text { and } y=k \end{aligned}

Where ‘ k ’ is a real number satisfy the given system of equations.

Posted by

Gurleen Kaur

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